Nonatomic measure space over set larger than the reals Question: Does anybody know a non-trivial nonatomic measure space over a set larger of cardinality larger than the reals? By non-trivial I mean that no set exists of cardinality equal to that of the reals, such that the complement is a null set. 
Background: I try to find a big list of examples of measure spaces, but I find these hard to find. Many examples either have something to do with the Lebesgue measure or are very complicated, so know I try to think of some measure spaces myself. 
Thank you in advance!
 A: If $S$ is any set, there is a measure $\mu$ on the set $X=\{0,1\}^S$ such that for each finite $F\subseteq S$ and each $a\in\{0,1\}^F$, $\mu(\{x\in X:x|_F=a\})=2^{-|F|}$ (this measure can be defined on the $\sigma$-algebra generated by these sets $\{x\in X:x|_F=a\}$ using the Caratheodory extension theorem, for instance).  Intuitively, you can think of this as the probability measure associated to independently flipping a coin for each element of $S$.  This measure is atomless as long as $S$ is infinite.  When $S$ is countably infinite, this is essentially Lebesgue measure on $[0,1]$ (think of the $0$s and $1$s as binary digits of a number).  But $S$ can be arbitrarily large, and if $S$ is sufficiently large then this $X$ will definitely be "nontrivial" in your sense.
(To prove the last claim, there are always at least $|S|$ different measurable subsets of $X$ mod null sets, so if $|S|>2^{2^{\aleph_0}}$, then $\mu$ cannot be concentrated on a set of size $2^{\aleph_0}$.  Probably actually $|S|\geq 2^{\aleph_0}$ suffices, but I don't see how to prove this.)
